100 Path Sum

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

  5
 / 4   8

/ / 11 13 4
/ ? 7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

S1：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
       
        if(!root) return false;
        if(!root->right && !root->left && root->val == sum) return true; 
        return(hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val));
        
    }
};

100 Path Sum

原文：https://www.cnblogs.com/liez/p/11634706.html